空白PIC12F1822的电流消耗以及60V的低功耗5V电源

      以下为原文

   

为什么不使用一个低静态，电流高效率专用调节器？TI有工具可以为你做大部分工作。



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    Why not use a low quiescent, current high Efficiency dedicated regulator? TI has tools that will do most of the work for you.

我确实找到了一些可以做的工作：HTTPS/2/256/MAX5033-1045 28PDF 76VIN 7VUA，70UA静态10UA关机3.50HTTPS://www. Mous.com／DS/2/256/Max 1562-257927.PDF 60Vin 5VUT 65%效率在5毫安48 VIN $2.50HTTP://www. Ti.COM/LIT/DS/Simulk/LM5165-Q1.PDF 6VIN，5VUT，205 UA有源，5UA关断，75%的效率在5毫安3.25美元，能够获得模拟小于1毫安从60V轨道抽出5V 5mA输出通过使用15mH电感器和2 kHz PWM与12USEC占空比。只有50%的效率。我还必须提供一些读取输出电压来调节PWM的方法。也许最好的总效率是在突发中运行调节器来充电一个电容器，它可以在样品之间的睡眠模式下为电路供电。这个BMS将被用于12V，12 AH的SLA电池，所以1Ma CON。在1000小时或超过一个月内，抽签将消耗10%的电荷。这可能类似于自我放逐。谢谢。



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    I did find some that might do the job:
 
https://www.mouser.com/ds/2/256/MAX5033-104528.pdf 76Vin 5Vout 270uA quiescent 10uA shutdown $3.50
 
https://www.mouser.com/ds/2/256/MAX15062-257927.pdf 60Vin 5Vout 65% efficiency at 5 mA out 48Vin $2.50
 
http://www.ti.com/lit/ds/symlink/lm5165-q1.pdf 65Vin, 5Vout, 205 uA active, 5uA shutdown, 75% efficiency at 5 mA out $3.25
 
I was able to get a simulation with less than 1 mA draw from the 60V rail for 5V 5mA output by using a 15mH inductor and 2 kHz PWM with 12uSec duty cycle. Only 50% efficiency. I'd also have to provide some means of reading the output voltage to adjust the PWM.
 
Perhaps best overall efficiency would be to run the regulator in bursts to charge a capacitor that can power the circuit in sleep mode between samples.
 
This BMS will be used on 12V, 12 A-h SLA batteries, so a 1mA constant draw would drain 10% of charge in 1000 hours, or over a month. That's probably similar to self-discharge.
 
Thanks.

那些调节器是线性的还是开关式的？我的研究将是使用汽车规格开关调节器从60V下降到5V。线性技术有一些，这是一个插头的情况下，他们给出的公式，并使用他们建议的电路板布局，你将有一个可靠的调节器与最少的组件。你应该能够获得比你引用的更高的效率。也应该有一些国家半导体从TI（具有LM零件号的）中获得，具有相似的能力和效率。



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    Are those regulators linear ones or switchmode ones?
 
My take on this would be to use an automotive specification switching regulator to drop from 60V to 5V. Linear Technology have some and it is a case of plug numbers onto the formula they give, and use their suggested board layout and you will have a reliable regulator with minimal components. You should be able to get much higher efficiency than you are quoting above.
 
There should also be some National Semiconductor available from TI (the ones with LM part numbers) with similar ability and efficiency. 

如果你去TIS网站，他们有一个选择工具。这将使你也参与到基于Web的设计和仿真工具中。你可能会发现周姑妈是一个更好的对手。一个中断模式PWM调节器可能是一个不错的选择。



      以下为原文

    I f you go to TIs web site they have a selection tool. That will take you in too their web based design and simulation tools. You may be able to find aunt tweek a better match. A discontinued mode pwm regulator may be a good choice.

我可能会使用MX5033。它是在SOIC-8和PDIP8封装，这将是更容易焊接的原型。其他的是以0.5毫米引线间距的2mm～3mm方形封装。把IC放在这样一个小封装中似乎是疯狂的，当辅助组件是10倍大时，这些都是开关稳压器，并且大多数都使用谐振ZV PWM开关技术以及脉冲跳变（PSM）用于轻负载。我不知道他们是如何能够达到这样的低静态电流和高效率，但可能有许多组件，可以在芯片中使用，当包装作为分立设备和安装在PCB上时受到影响。我想我会继续做一个自己设计的样机，只是为了了解更多关于高效率微功率开关稳压器的知识。如果我需要特殊应用的东西，比如EV电池组或线电压（高达350 VDC）的高输入电压，可能会有用。另一种方法是使用光隔离器作为控制元件，所以对高电平没有任何实际限制。除PMOS晶体管外的IDE总线电压。大部分的功耗来自电阻连接从低5VDC电源到高侧开关元件，高电阻值慢下来的事情，因为电容。我选择了具有Schmitt trigger的H11L1，它具有2的USEC响应时间和100n秒的上升/下降时间来驱动PMOS门。LED驱动至少5Ma，但占空比小于1%，平均为50uA。但它也需要一个高端的稳压电源，这就是为什么我必须拿出一个高效的调节器。此外，低侧PIC在驱动高侧开关电路之前需要进行供电。我喜欢使用白色LED作为分路调节器的线性稳压器的想法。两个串联在12uA处提供140UA和4.88伏的5.13VDC，但是PIC似乎将它们加载到仅获得2.35V。然而，添加晶体管提供了一个实心4.5VDC，而单个LED将为低功率器件提供1.8VDC。



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    I'll probably use the MAX5033. It comes in an SOIC-8 and PDIP-8 package which will be a lot easier to solder on the prototype. The others come in tiny 2mm-3mm square packages with 0.5mm lead spacing. It seems crazy to put an IC in such a tiny package when the ancillary components are 10 times bigger.
 
These are all switching regulators, and most of them use resonant ZV PWM switching techniques along with pulse-skipping (PSM) for light loads. I wonder just how they are able to achieve such low quiescent current and high efficiency, but there are probably many components that can be utilized in a chip, that suffer when packaged as discrete devices and mounted on a PCB. I think I will continue to make a prototype of my own design just to learn a bit more about high efficiency micropower switching regulators. Might come in handy if I need something for special applications, such as perhaps higher input voltages as seen in EV battery packs or line voltage (up to 350 VDC).
 
Another method I considered was to use an opto-isolator as the control element, so there would be no practical limitation to the high side bus voltage except for the PMOS transistor. Much of the power dissipation comes from resistive connections from the lower 5VDC supply to the high-side switching components, and high values of resistance slow things down because of capacitance. I chose an H11L1 with Schmitt trigger which has 2 uSec response time and 100nSec rise/fall time to drive the PMOS gate. The LED must be driven with at least 5mA but since the duty cycle is less than 1% it is an average of just 50uA. But it also needs a regulated power supply on the high side, and that's why I had to come up with an efficient regulator. Also, the low-side PIC would need to powered before it could drive the high side switching circuit.
 
I do like the idea of a linear regulator using white LEDs as shunt regulators. Two in series provides 5.13VDC at 140uA and 4.88V at 12uA, but the PIC seems to load them down to get only 2.35V. However, adding a transistor provides a solid 4.5VDC, and a single LED would provide 1.8VDC for low power devices.

我编程了PIC的最低功率（我认为），但仍然只有2.35 VDC上的VDD和VSS的设备，与LED和249K电阻到40伏直流电。我认为它应该只画出20个UA典型和40个UA最大值，这对应于一个5/40＝125K的电阻，所以它将是大约13伏作为分压器。但是，LED可能具有低于阈值的显著泄漏电流。它们在12 UA上下降4.88伏，相当于406K，在140 UA，5.13V，这是128K。这是我的代码（借用另一个项目）：



      以下为原文

    I programmed the PIC for lowest power (I think), but there was still only 2.35 VDC on the Vdd and Vss of the device, with the LEDs and a 249k resistor to 40 VDC. I would think it should draw only 20 uA typical and 40 uA maximum, which corresponds to a resistor of 5/40 = 125k so that would be about 13 volts as a voltage divider. But perhaps the LEDs have a significant leakage current below threshold. They drop 4.88V at 12 uA which is equivalent to 406k, and 5.13V at 140 uA, which is 128k.
 
Here is my code (borrowed from another project):
// CONFIG1
#pragma config FOSC = INTOSC    // Oscillator Selection (INTOSC oscillator: I/O function on CLKIN pin)
#pragma config WDTE = OFF       // Watchdog Timer Enable (WDT disabled)
#pragma config PWRTE = OFF       // Power-up Timer Enable (PWRT enabled)
#pragma config MCLRE = OFF      // MCLR Pin Function Select (MCLR/VPP pin function is digital input)
#pragma config CP = OFF         // Flash Program Memory Code Protection (Program memory code protection is disabled)
#pragma config CPD = OFF        // Data Memory Code Protection (Data memory code protection is disabled)
#pragma config BOREN = OFF       // Brown-out Reset Enable (Brown-out Reset enabled)
#pragma config CLKOUTEN = OFF   // Clock Out Enable (CLKOUT function is disabled. I/O or oscillator function on the CLKOUT pin)
#pragma config IESO = ON        // Internal/External Switchover (Internal/External Switchover mode is enabled)
#pragma config FCMEN = OFF      // Fail-Safe Clock Monitor Enable (Fail-Safe Clock Monitor is disabled)

// CONFIG2
#pragma config WRT = OFF        // Flash Memory Self-Write Protection (Write protection off)
#pragma config PLLEN = OFF      // PLL Enable (4x PLL disabled)
#pragma config STVREN = ON      // Stack Overflow/Underflow Reset Enable (Stack Overflow or Underflow will cause a Reset)
#pragma config BORV = LO        // Brown-out Reset Voltage Selection (Brown-out Reset Voltage (Vbor), low trip point selected.)
#pragma config LVP = OFF        // Low-Voltage Programming Enable (High-voltage on MCLR/VPP must be used for programming)

// #pragma config statements should precede project file includes.
// Use project enums instead of #define for ON and OFF.

#include

#include
#include

#define DS3         LATAbits.LATA5  //pin2 GP5
#define DS2         LATAbits.LATA2  //pin5 GP2
#define TAP         ANSA4           //pin3 GP4
#define _XTAL_FREQ  4000000
#define ERROR       51              // 3.2V/20=160 mV
/*
 *
 */
void main(void) {
    int     ADCvalue = 0;
    
    OSCCON = 0b00000000;    // 32 kHz LF
    TRISA = 0;  // 0b00011011;     //GP2, GP5 output
//    ANSELA = 0b00010000;    //AN3 GP4
    LATA = 0b00000000;
//    ADCON0 = 0b00001101;    // AN3 ADC ON
//    ADCON1 = 0b10001000;    // Right Just, Osc/8, Vref=Vdd
//    __delay_ms(1000);
    while (1) {
        sleep();
        ADCON0bits.ADGO=1;
        while(ADCON0bits.nDONE);
        ADCvalue = 256*ADRESH+ADRESL;
        if(ADCvalue > 512+ERROR) {
            LATAbits.LATA2 = 1;
            __delay_ms(50);            
            }
        else if(ADCvalue < 512-ERROR) {
            LATAbits.LATA5 = 1;
            __delay_ms(50);
            }
        LATA = 0;
        __delay_ms(1000);
    }
        
}

对这条线的评论是错误的，如果你启用BOR有什么区别吗？



      以下为原文

    The comment on this line is wrong.
Does it make any difference if you DO enable BOR?
#pragma config BOREN = OFF       // Brown-out Reset Enable (Brown-out Reset enabled)
 

评论没有改变以前的项目。我只是尝试了一种不同类型的设备，它也显示了只有2.35伏特的电源引脚。如果我分别测量每个LED，我读0.97和0.60伏特，显示万用表的效果。这似乎表明，LED电流是最小的，可能小于1 UA。[编辑] I测量设备的电流牵引与5VDC电源，它是925 UA，在2.5VDC，它是720 UA。[编辑2]我尝试了一个空白PIC16F1825，它画出300 UA在5伏和150 UA在2.5伏。



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    The comments were not changed from the previous project. I just tried a different device of same type, and it also shows only 2.35 volts across the power pins. If I measure each LED separately, I read 0.97 and 0.60 volts, showing the effect of the multimeter. That seems to indicate that the LED current is minimal, probably less than 1 uA.
 
[edit] I measured current draw of the device with a 5VDC supply and it is 925 uA, and at 2.5VDC it's 720 uA.
 
[edit2] I tried a blank PIC16F1825 and it draws 300 uA at 5 volts and 150 uA at 2.5 volts.

哎呀！我发现我的项目中有一个不同的主文件.C文件，所以我的修改不适用。修复后，PIC12F1822在5V处只吸引68 UA（其中LED几乎不发光，所以绘制电流），而在4 VDC，电流仅为49 UA。采用并联稳压器和249K电阻串联LED，可吸收4.12毫安。LED发光适度明亮，并调节VDD至5.12伏直流电。这是34.8伏直流跨越249K电阻，所以实际电流应为1.40毫安。我的超级Duffer实验室级港口货运万用表测量电阻为205K在一个方向和260K在另一个。我想我需要清理我的PCB！清洗PCB产生了奇迹。电阻器现在读取245-250K。PIC绘制65伏在5伏和39 UA在4伏。此外，我忘了我有一个10K电阻跨越电路的输入，它本身在40伏抽出4毫安。电路本身绘制200 UA在40 VDC.YABBA-DABB-DOO！



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    OOPS! I just found out that I had a different main.c file in my project, so my changes did not apply. After fixing that, the PIC12F1822 draws just 68 uA at 5V (where the LEDs barely glowed, so drawing current), and at 4 VDC the current is just 49 uA. With the shunt regulator and the 249k resistor in series with the LEDs, it draws 4.12 mA. The LEDs glow moderately bright, and regulate Vdd to 5.12 VDC. That's 34.8 VDC across the 249k resistor, so actual current should be 1.40 mA. My super duper lab grade Harbor Freight free multimeter measures the resistor as 205k in one direction and 260k in the other. I think I need to clean my PCB!
 
Cleaning the PCB did wonders. The resistor now reads 245-250k. The PIC draws 65 uA at 5 volts and 39 uA at 4 volts. Also, I forgot that I have a 10k resistor across the input of the circuit, which itself draws 4 mA at 40 volts. The circuit itself draws 200 uA at 40 VDC.
 
Yabba-Dabba-Doo!

为什么使用这个拓扑？VS控制器在低侧，由输出供电。在第二种情况下，分流可以在启动后关闭（低静态电流），并且接地对于所有电路都是通用的。艾伯特



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    Why you use this topology? vs  controller on low side, powered from output.
In second case shunt can be turned off after startup(lower quiescent current)
and ground is common for all circuits.
 
Albert

将有一个控制器在低侧执行BMS功能，但我需要提供5 VDC电源从高侧总线，这可能是从12 VDC到60 VDC的任何东西。在5V（25MW）的5Ma线性调节器将从高母线中抽出至少5Ma（300MW），这是不希望的。降压开关电源应该能够做到这一点，只要在60V（60MW 42%效率）或2.5Ma在12V（30MW 83%的效率）为1Ma。如上所述，有专用的开关控制器将满足这一规范。通过使开关电源每1秒采样仅100MSEC，有效电流消耗可以减少到1/10。因此，即使是线性供应也可以平均500 UA包抽签。两个白色LED可以在50uA上连续地保持，它们将显著发光并提供良好的5V参考。低端控制器可以禁用电源的缓冲输出，但在休眠模式下，它只会吸引大约50 UA。我可能会继续基于这个设计制作一个原型。一些电路和代码对于各种用途可能是有用的，即使专用开关电源对于这个特定的应用更实用。



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    There will be a controller on the low side to perform the BMS functions, but I need to provide 5 VDC power from the high side bus, which might be anything from 12 VDC to 60 VDC. A linear regulator for 5mA at 5V (25mW) would draw at least 5mA (300mW) from the high bus, which is undesirable. A buck switching supply should be able to do that with as little as 1mA at 60V (60mW 42% efficiency) or 2.5mA at 12V (30mW 83% efficiency). As noted above, there are dedicated switching controllers that will meet this specification.
 
By enabling the switching supply for only 100mSec for every 1 second sample, the effective current drain can be reduced to 1/10. So even a linear supply could function with an average of 500 uA pack draw. The two white LEDs could be left on continuously at 50uA at which they will glow noticeably and provide a good 5V reference. The low side controller could disable the buffered output of the supply, but with the device in sleep mode it will draw only about 50 uA more.
 
I will probably continue making a prototype based on this design. Some of the circuitry and code may be useful for various purposes, even if a dedicated switching supply is more practical for this specific application.

这里是一个简单的线性电源的仿真，它可以工作在30V到60V的包电压（或12V到130V，对于R1的值不同）。它假定一个低电平的PIC，每秒大约100毫秒工作在5Ma左右。电源电压从4.12到4.69伏变化，平均电流低于1Ma：



      以下为原文

    Here is a simulation of a simple linear power supply that can work on 30V to 60V pack voltage (or 12V to 130V with different values for R1). It assumes a PIC on the low side that operates at about 5mA for 100 mSec out of every second. The supply voltage varies from 4.12 to 4.69 volts and average current is well under 1mA:
 

 

好啊。所以你把控制MOSFET的难度转换成控制器的线性电源和缺乏共同点。你的选择，相当聪明。我试着比较这样一个解决方案的利润和损失，而不是破坏它。我为我的英语缺乏精确性道歉。RT



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    OK. So you traded the difficulty of controlling the upper MOSFET
for the controller's linear power supply and lack of common ground.
Your choice, quite clever.
I am trying to compare the profits and losses of such a solution, and not to undermine it.
I apologize for the lack of precision of my English.
 
Albert

我欣赏所有的评论，它们帮助我以不同的眼光看待设计。不要担心你的英语——这是相当充分的，我想我理解你的意思。它为我更全面的分析和更好的设计决策打开了大门。谢谢。[编辑]这里是一些电压调节器的实际数据点，在睡眠模式下为PIC12F1822供电。我使用一个MMBTA06晶体管，249K电阻，1μF电容器在PIC上，和两个白色LED串联，这明显地发光，只有6 VDC的应用：Vin Vout Iin（UA）6 4.35 21.51 2.0 2.0 4.50 44.020.0 4.57 76830.0 4.62 4.62 115 44 0.0 4.65 153.65 0.0 4.68 194.4



      以下为原文

    I appreciate all the comments, and they help me look differently at the design with an eye to improvement. Don't worry about your English - it is quite adequate and I think I understood what you meant. It opened the door to my more comprehensive analysis and better design decisions. Thanks.
 
[edit] Here are some actual data points for the voltage regulator, powering the PIC12F1822, in sleep mode. I am using an MMBTA06 transistor, 249k resistor, 1 uF capacitor across the PIC, and two white LEDs in series, which visibly glow with as little as 6 VDC applied:
 
Vin   Vout   Iin(uA)
 6.0  4.35   21.5
12.0  4.50   44.0
20.0  4.57   76.8
30.0  4.62  115.4
40.0  4.65  153.6
50.0  4.68  194.4
 

考虑使用耗尽模式MOSFET而不是R1（在α13）中的宽范围输入电压，如HTTPS://wwwimuneN.COM/DGDL/Unimeon Apple；文件ID＝546D4624CB7F111014CD63D1A197D94图5？Zener二极管也可以是LED或上侧，考虑艾伯特。



      以下为原文

    Consider to use depletion mode MOSFET instead R1(in #13)  for wide range input voltage as in
https://www.infineon.com/dgdl/Infineon-Application_Note_Applications_for_Depletion_MOSFETs-AN-v01_00-EN.pdf?fileId=5546d4624cb7f111014cd63d1a197d94
Figure 5 ?
Zener diode may be LED or upper side as well.
 
Regards Albert

你好，保罗，你的PS之所以低是因为PNP的基极电流有限。如果你想达到低成本，你可以搜索电路“使用晶体管的并联电压调节器”，但是你必须考虑晶体管的功耗。



      以下为原文

    Hi Paul,
The reason you have low PS is due to limited base current to the PNP. 
If you want to achieve low cost, you can search the circuits " shunt voltage regulator using transistor ", but you have to consider the power dissipation on transistor.

PIC不需要有5伏。它只有12V的10%低，并保持在0.18伏到50V以内。MMBTA06的beta值至少为100，所以当PIC和其他电路绘制5毫安时，它应该调节得足够好。最大功率为60×5＝300 mW，额定功率为350 mW。此外，10%的占空比平均只有30兆瓦，BSP179的数量是0.83美元，10，而MMBTA06是0.14美元。我并不需要一个400伏的设备，但如果我做了BSP179可能是一个不错的选择。MMBTA06额定为80V和500毫安，这是充足的，但感谢信息。我不太熟悉耗尽型MOSFET。它们类似于JFETs，像2N319，这是我多年前使用的（70年代初）。.



      以下为原文

    It is not necessary to have 5.0 volts for the PIC. It's only 10% low at 12V in and holds that to within 0.18 volts up to 50V. The MMBTA06 has a beta of at least 100 so it should regulate well enough when the PIC and other circuitry draw 5 mA. The maximum power will be 60*5 = 300 mW and it's rated for 350 mW. Besides, the 10% duty cycle results in just 30 mW average.
 
The BSP179 is $0.83 in quantity 10, while the MMBTA06 is $0.14.
 
I don't really need a 400V device, but if I did the BSP179 might be a good choice. The MMBTA06 is rated 80V and 500mA, which is plenty.
 
But thanks for the info. I was not very familiar with depletion mode MOSFETs. They are similar to JFETs, like the 2N3819, which I have used many years ago (early 1970s!).

你说得对，它们非常类似于JFETs，但却不太出名，所以我敢提出一个链接。你的计算和价格数据总是完全正确的。但是我并不意味着电压范围。从你的帖子16看数据，你只需要4UA（（6 -（4.35 + 0.7））V/249KOHM）流过LE。D，但你消耗180uA（（50 -（4.68 + 0.7））V/249KOHM）在50V。通过使用我建议的电流源，无论电源电压如何，都可以优化电流（例如10uA或其它）。检查价格BSS169、BSS159等或Microchip（SuthTeX）产品



      以下为原文

    You're right, they're very similar to JFETs, but less known, so I dared to put a link.
Your calculations and price data are completely correct, as always.
But I did not mean voltage ranges.
Looking at the data from your post # 16 you only need 4uA ((6- (4.35 + 0.7)) V / 249kohm) flowing through the LED, but you consume 180uA ((50- (4.68 + 0.7)) V / 249kohm) at 50V.
By using the current source that I suggested, you can set the current optimally (eg 10uA or whatever) regardless of the power supply voltage.
I am also interested in how this power source will behave when the output MOSFET capacity will be charged/discharged.
 
PS.
Check price BSS169, BSS159 etc. or Microchip (Supertex) products
 
Albert